## Ivycomptech Interview Question for Software Engineer / Developers

Country: United States
Interview Type: In-Person

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1
of 1 vote

``````int main() {
int a[20],n,k;
scanf("%d",&n);
for(int i =0 ;i<n;i++) {
scanf("%d",&a[i]);
}
int c = 0;
for ( int i = 0; i< n;i++) {
for (int j = i+1;j<n;j++) {
if(a[i]>a[j]) {
printf("%d %d\n",a[i],a[j]);
c++;
}
}``````

}

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1
of 1 vote

Modifed version of mergesort algorithm could give us the no of inversions in the array. I tested this on
(1), (1,1,1,1), (1,2,3,4,5,6), (3,5,7,2,8)

``````package arrays;

public class FindIInversionCount {

public static int[] temp_data;
public static int inversion_count = 0;

public static void main(String[] args) {
int[] a = {3,5,7,2,8};
temp_data = new int[a.length];
inversion_count = merge_and_count(a, 0, a.length - 1);

//		for (int i : temp_data) {
//			System.out.println(i);
//
//		}
System.out.println("inversion count: " + inversion_count);
}

private static int merge_and_count(int[] a, int low, int high) {
if (high - low > 1) {
// more than two element

int mid = (int) Math.floor((high + low) / 2);

merge_and_count(a, low, mid);
merge_and_count(a, mid + 1, high);

// merge the data and count the inversions
int i = low;
int j = mid + 1;
int count = low;
while (i <= mid && j <= high) {
if (a[i] <= a[j]) {
temp_data[count++] = a[i];
i++;
} else if (a[i] > a[j]) {
temp_data[count++] = a[j];
j++;
for (int k = i; k <= mid; k++) {
inversion_count++;
}
}
}

if (j <= high) {
while (j <= high) {
temp_data[count++] = a[j];
j++;
}
}
if (i <= mid) {
while (i <= mid) {
temp_data[count++] = a[i];
i++;
inversion_count++;
}
}
} else {
// we have to deal with less than two elements
if (high == low + 1) {
if (a[low] > a[high]) {
int tmp = a[low];
a[low] = a[high];
a[high] = tmp;
inversion_count++;
}
} else if (high == low) {
System.out.println("One element no sorting needed");
temp_data[low] = a[low];
}
}
return inversion_count;
}
}``````

Comment hidden because of low score. Click to expand.
1
of 1 vote

I think I have solution.
I am not sure about running time as I haven't calculated it yet.

Let's say we have an array of 11 elements (10,14,12,18,7,9,13,2,6,15,1)
Possible inversions :
(10,8) (10,7) (10,9) (10,2) (10,6) (10,1)
(14,12) (14,8) (14,7) (14,9) (14,13) (14,2) (14,6) (14,1)
(12,8) (12,7) (12,9) (12,2) (12,6) (12,1)
(8,7) (8,2) (8,6) (8,1)
(7,2) (7,6) (7,1)
(9,2) (9,6) (9,1)
(13,2) (13,6) (13,1)
(2,1)
(6,1)
(15,1)

Total : 36 Inversions

Now solution :

1. Let's define element with their positions.

10-1
14-2
12-3
8-4
7-5
9-6-
13-7
2-8
6-9
15-10
1-11

2. Find Minimum element , in the above case we have 1 at position 11 , so all above 1 will be greater than 1 -- 10 Inversion.

3. Find next minimum element and exclude 1 , in above case it's 2 at position 8 -- 7 Inversion.

4. Repeat step 3, next minimum element 6 at position 9 , Now exclude 2 (as it already been processed - Inversion (9 (position of 6 ) -1 (2 is already processed) - 1 (as element 6 itself )
--7 Inversion.

5. Repeat step 3 & 4 and in each iteration we will get number of inversions.

6. At last add all inversions--10+7+7+4+3+3+0+1+1+0+0 = 36 Inversion total.

Let me know if it doesn't apply to all case.
I will try to optimize it and write the running time.

Thanks,

Pankaj T.

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0
of 0 vote

Sorry, the above output was not mentioned correctly.
The Output should be 3. ( It has 3 inversions: 3>2,, 5>2, 7>2)

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0
of 0 vote

I forgot to mention the length of the array is unknown and your code should be optimized in less than O(n^2).

Comment hidden because of low score. Click to expand.
0
of 2 vote

number of inversions = sum of distance of (current position of each element, its position in sorted array) / 2.
Prove by induction. n = 1 case is trial.
n = k to n = k+1: suppose the largest element is inserted to position i.
elements behind it increase the distance by 1, there are k+1-i of them;
itself has the distance of k+1-i.
This insertion created k+1-i new inversions.
So the conclusion is valid for n=k+1.

Back to the question, now it's easy to learn it has a solution of O(n*log(n)) with sorting and then counting distances.

Any idea on an O(n) solution?

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0

Hi cx9,

I have mentioned the condition i < j of the given array. If you sort the array, the original positions (i, j) will differ.

In the eg(given in question), there are only 3 inversions.

If you sort the array then the number of inversions will become n(n+1) / 2. = 5 (5+1) / 2. = 15 which is wrong answer.

Comment hidden because of low score. Click to expand.
0

Maybe you don't understand what I mean.
In your example: {3, 5, 7, 2, 8}
The positions of elements are:
p(3) = 1, p(5) = 2, p(7) = 3, p(2) = 4, p(8) = 5;
in the sorted array {2, 3, 5, 7, 8}
p'(2) = 1, p'(3) = 2, p'(5) = 3, p'(7) = 4, p'(8) = 5;
And the distance I talked about is such as:
|p(3)-p'(3)| = |1-2| = 1.
So the sum is |1-2| + |2-3| + |3-4| + |4-1| + |5-5| = 6;
Output is 6/2 = 3.

My algo needs O(n) extra space.

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0

``````Ok cx9,
I understood your logic. But when I tried for below array, it's not getting expected answer.
arr = {5,1,6,4,3,9}
Expected Output: 6  [ (5,1),  5,4), 5,3), (6,4), (6,3), (4,3)]

sortedarr = {1,3,4,5,6,9}

Sum of their distances = (|4-1| + |2-1| + |3-5| + |4-3| + |5-2| + |6-6|) / 2  =  10 / 2  =  5 which is incorrect number of inversions.

Please correct me  if I missed any steps.``````

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0

Yes, Veeru you are right. There is logical incorrectness in my algo. This statement in my first reply "suppose the largest element is inserted to position i. elements behind it increase the distance by 1elements behind it increase the distance by 1" doesn't hold. It's easy to give a counter example. So my algo is a wrong one.

Comment hidden because of low score. Click to expand.
0
of 0 vote

divide and conquer, similar to merge sort. O(nlgn)

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0
of 0 vote

For each element in an array
--->Insert into a BST and count the number of successors in the BST.
the no if successors is the no of inversions for that element.
Since we need the count of the successors we can increase the efficiency by maintaining the number of nodes in right subtree at each node.
This should be a nlogn .

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